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- how can we use the Schrodinger equation while finding ##\hat p## when in fact we have already used ##\hat p##(i.e. ##\hat p ^2## in the ##\hat T## term of the ##\hat H##) in the Schrodinger equation?

I have read that the Schrodinger equation has no formal derivation we are simply applying the Hamiltonian operator on the wave function

$$\hat H = i\hbar \frac{\partial}{\partial t} = \hat T + \hat V$$

here we substitute $$\hat T = \frac{\hat p^2}{2m}$$ where $$\hat p = -i \hbar \frac{\partial}{\partial x}$$

but when we derive the equation for ##\hat p## we actually substitute ##\frac{\partial \psi}{\partial t}## and ##\frac{\partial \psi *}{\partial t}## from the Schrodinger equation.

$$< p> = m\frac{d<x>}{dt} = m\int_{-\infty}^{+\infty} x\frac{\partial (\psi^*\psi)}{\partial t}$$

$$<p> = m\int_{-\infty}^{+\infty} x[\psi^*\frac{\partial \psi}{\partial t}+\psi\frac{\partial \psi^*}{\partial t}] dx$$

here we substitute ##\frac{\partial \psi}{\partial t}## and ##\frac{\partial \psi *}{\partial t}## as

$$\frac{\partial \psi}{\partial t} = \frac{i\hbar}{2m}\frac{\partial^2 \psi}{\partial x^2} -\frac{i}{\hbar} V\psi$$

$$\frac{\partial \psi^*}{\partial t} = \frac{-i\hbar}{2m}\frac{\partial^2 \psi^*}{\partial x^2} +\frac{i}{\hbar} V\psi^*$$

after some simplification we end up with

$$<p> = \int_{-\infty}^{+\infty} \psi^* (-i\hbar \frac{\partial}{\partial x})\psi dx$$

and then finally we get $$\hat p = -i\hbar \frac{\partial }{\partial x}$$

so I don't understand how we can use the Schrodinger equation while finding ##\hat p## when in fact we have already used ##\hat p##(i.e. ##\hat p ^2## in the ##\hat T## term of the ##\hat H##) in the Schrodinger equation?

this video does the derivation for the momentum operator

$$\hat H = i\hbar \frac{\partial}{\partial t} = \hat T + \hat V$$

here we substitute $$\hat T = \frac{\hat p^2}{2m}$$ where $$\hat p = -i \hbar \frac{\partial}{\partial x}$$

but when we derive the equation for ##\hat p## we actually substitute ##\frac{\partial \psi}{\partial t}## and ##\frac{\partial \psi *}{\partial t}## from the Schrodinger equation.

$$< p> = m\frac{d<x>}{dt} = m\int_{-\infty}^{+\infty} x\frac{\partial (\psi^*\psi)}{\partial t}$$

$$<p> = m\int_{-\infty}^{+\infty} x[\psi^*\frac{\partial \psi}{\partial t}+\psi\frac{\partial \psi^*}{\partial t}] dx$$

here we substitute ##\frac{\partial \psi}{\partial t}## and ##\frac{\partial \psi *}{\partial t}## as

$$\frac{\partial \psi}{\partial t} = \frac{i\hbar}{2m}\frac{\partial^2 \psi}{\partial x^2} -\frac{i}{\hbar} V\psi$$

$$\frac{\partial \psi^*}{\partial t} = \frac{-i\hbar}{2m}\frac{\partial^2 \psi^*}{\partial x^2} +\frac{i}{\hbar} V\psi^*$$

after some simplification we end up with

$$<p> = \int_{-\infty}^{+\infty} \psi^* (-i\hbar \frac{\partial}{\partial x})\psi dx$$

and then finally we get $$\hat p = -i\hbar \frac{\partial }{\partial x}$$

so I don't understand how we can use the Schrodinger equation while finding ##\hat p## when in fact we have already used ##\hat p##(i.e. ##\hat p ^2## in the ##\hat T## term of the ##\hat H##) in the Schrodinger equation?

this video does the derivation for the momentum operator

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